Main     Puzzle 8M85 by Richard Pavlicek    

Cold Slam in Hand

“Professor! You won’t believe this monster I picked up with Grover last night. My right-hand opponent opened 3 D, and I almost fell out of my chair. Look at this mitt! A cold grand in my own hand!”

S A K Q J 10
H A K Q J 9 8 7 6

“Nice hand, Timothy, but it’s hardly cold for seven. Five hearts in one hand, or an opening spade ruff would put you down. In fact, there are only four hands out of 635+ billion that are cold for seven in a suit. It takes a flush hand; anything less, and something could go wrong.”

“I suppose, but who worries about that. I thought about bidding it deceptively, but nothing was safe with Grover (he even passes cue-bids) so I jumped to 7 H. It was cold but a tie for bottom, since everyone else did the same, and five tables were doubled and redoubled. My left-hand opponent smartly passed with two aces. I guess I had to hope for one of those fluke distributions you say to get a tie for top. Strange game, duplicate. At least I was cold for a small slam.”

“Sorry to disillusion you, Timothy, but you can’t even claim six. However unlikely, though consistent with my boson deflection theory, 6 H could be set by a spade ruff with a singleton trump, then a diamond to promote a trump trick for left-hand opponent who is also void in diamonds. You needed the H 10 to be cold for six.”

“Whatever. Are you suggesting I shouldn’t even bid a slam?”

“No, but you have to gauge your bidding to bosonic theory. A successful contract is not a simple wave but a complex collision of subnuclear particles. Slam bidding especially requires a strategic approach, not a surge to the verge. Off hand, I don’t recall the exact number of hands cold for six of a suit, but it’s fewer than those cold for 6 NT. Directly confronting this threshold is to deny the mere existence of bosons and could set bridge theory back 30 years.”

“Professor? Get a life! Every time I bring up a bridge hand you turn it into a lecture on some bozo theory of yours. Does anybody really care how many hands there are?”

[Richard raises hand]. In case there’s anybody else out there, this will be the June puzzle:

How many bridge hands are guaranteed to win 12 tricks in a suit?

For the sake of argument, do not count the four hands cold for 13 tricks. As a tiebreaker, include your best guess to the related question: How many hands are guaranteed to win 12 tricks (but not 13) in notrump? “Guaranteed” means you must succeed against any distribution, any defense, and without having to guess anything about the layout.

Try it now

Make your best guesses and click Verify to find out what, if anything, is wrong. Use the help provided to make corrections and repeat. See how many tries it takes you to find the answers. Both answers are 5-digit numbers.

How many bridge hands are guaranteed
to win 12 tricks in a suit?

Your answer
How many bridge hands are guaranteed
to win 12 tricks in notrump?
Your answer
Do not count hands cold for 13 tricks


Dan Baker Wins!

In June 2016 this puzzle was presented as a challenge — with no help provided — inviting anyone who wished to submit a solution. Participation was below the previous month’s high mark but still third most in the series, with 55 persons giving it a try, of which only the 15 listed found the correct answer to the main question. (I decided to cut the listing there since none of the misses were close.)

Don’t mess with Texas! Congratulations to Dan Baker, who was the first of only four solvers to come up with the correct answer to the more difficult tiebreaker. This is Dan’s first win, though he’s been a steady contender, populating most of my leaderboards. The other three perfect solvers were certainly no surprise, as adding up their combined wins might be as complicated as the problem.

RankNameLocation  Answer  Tiebreaker
1Dan BakerTexas4734048360
2Dan GheorghiuBritish Columbia4734048360
3Tim BroekenNetherlands4734048360
4Charles BlairIllinois4734048360
5Julien ReichertFrance4734048300
6Ivan LoyHong Kong4734048444
7Gareth BirdsallEngland4734048260
8Rob DixonEngland4734048260
9Grant PeacockMaryland4734048252
10Jamie PearsonOntario4734048600
11Dominique CannevaFrance4734047940
12Wayne SomervilleNorthern Ireland4734047808
13Jean-Christophe ClementFrance4734047572
14Sherman YuenSingapore4734047424
15Tom SlaterEngland4734050532



For the main question, one fact should be clear: Any hand cold for 12 tricks in a suit must have two voids, as otherwise it would be an easy setup for the defenders to crossruff two side suits. With that in mind, listing the possible hand types is straightforward, as shown in the table below. A question mark (?) means any card below the lowest specified card (or in Case 1 any card, period); the ‘x’ in Case 2 means any card but the ace. All that remains is to determine the number of combinations for each case and multiply by the number of ways, which is always 12, since any hand with two voids has 12 suit permutations. Adding up the hands for each case gives the answer of 47,340.

CaseHand TypeCombinationsWaysHands
1???????????? A13c1212156
2AK?????????? x11c10 × 12c1121584
3AK????????? A?11c9 × 12c1127920
4AKQ??????? AK?10c7 × 11c11215840
5AKQJ????? AKQ?9c5 × 10c11215120
6AKQJT??? AKQJ?8c3 × 9c1126048
7AKQJT9? AKQJT?7c1 × 8c112672
Total hands guaranteed to win 12 tricks in a suit47340

With 12 trumps and a side ace (Case 1) trump quality makes no difference, since the most you could ever lose is one trump trick; but with a side loser (Case 2) they must be headed by A-K to withstand a potential trick-one loss and trump promotion. In Cases 3-7, trumps need only be strong enough to ensure drawing the missing trumps, since an opening ruff spends a trump, allowing you to ruff high on the return. Also note that the side suit does not have to be solid like Timothy’s A-K-Q-J-10; the lowest card can be a loser, since it vanishes with an opening ruff, and otherwise will be the only trick lost.

Dan Baker: The hand must have two voids (11-1-1-0 could suffer a crossruff); 7-6 must be AKQJT9? AKQJT? (7 × 8), [Dan accounts similarly for all other shapes]; then multiply by 12 for suit identity.

Jamie Pearson: Declarer must have exactly two suits (a third could lead to two ruffs). There must be enough top trumps not to allow an opening ruff and trump promotion, and at most one non-winner in the other suit. From there it’s just crunching the numbers and dealing with the special case of 12 trumps.

Sherman Yuen: Declarer must have exactly one side suit. If declarer has 7-11 trumps, the defense can always start with a ruff, followed by leading declarer’s void for a trump promotion; so the minimum holding is the top N trumps and the top N-1 cards in the side suit, where N is the number of missing trumps. If declarer has 12 trumps, they can be headed by AK with any singleton, or AQ/KQ with a singleton ace.

Gareth Birdsall: The hand must have only two suits, else concede a crossruff, and the lowest card in the side suit may be a loser. Total is: 12 × (2 + 11×13 + 11c9×12 + 10c7×11 + 9c5×10 + 8c3×9 + 7c1×8). [Gareth accounts for my Cases 1-2 differently, as did Sherman above, but it’s just as correct].

Dan Gheorghiu: A straightforward calculation. The hand cannot contain more than two suits for the reason explained in the preamble. Various lengths of the two suits, starting with 7-6 and ending with 12-1, yield a bell-shaped distribution, which peaks at 9-4 and 10-3 with almost the same number of hands.

Notrump is made of sterner stuff

The tiebreaker for hands guaranteed to make 6 NT was more difficult, as many more possibilities exist. Besides hands with 12 top tricks, one must account for hands able to establish a twelfth trick. My approach was to break it down into 51 cases listed in the following table, which is ordered by the length, secondarily by the strength, of the suit with a potential loser. Notations used: H = any card above the highest specified card; ? = any card below the lowest specified card; x = any card below and not touching the lowest specified card. Tinted rows indicate cases where the twelfth trick must be established, mandating at least two cards (A-K) in each solid suit.

CaseHand TypeCombinationsWaysHands
1AKQJTxx AK AK AK7c2484
2HHHH98? AK AK AK5c4 × 6c14120
3AKQJTx AKQJT A A7c11284
4AKQJTx AKQJ AK A7c124168
5AKQJTx AKQ AKQ A7c11284
6AKQJTx AKQ AK AK7c11284
7HHHH98 AKQ AK AK5c41260
8AKQJx AKQJT9 A A8c11296
9AKQJx AKQJT AK A8c124192
10AKQJx AKQJ AKQ A8c124192
11AKQJx AKQJ AK AK8c11296
12AKQJx AKQ AKQ AK8c11296
13HHHT9 AKQJ AK AK4c31248
14HHHT9 AKQ AKQ AK4c31248
15AKQx AKQJT9? A A9c1 × 7c112756
16AKQx AKQJT9 AK A9c124216
17AKQx AKQJT AKQ A9c124216
18AKQx AKQJT AK AK9c112108
19AKQx AKQJ AKQJ A9c112108
20AKQx AKQJ AKQ AK9c124216
21AKQx AKQ AKQ AKQ9c1436
25AKx AKQJT??? A A10c1 × 8c3126720
26AKx AKQJT9? AK A10c1 × 7c1241680
27AKx AKQJT9 AKQ A10c124240
28AKx AKQJT9 AK AK10c112120
29AKx AKQJT AKQJ A10c124240
30AKx AKQJT AKQ AK10c124240
31AKx AKQJ AKQJ AK10c112120
32AKx AKQJ AKQ AKQ10c112120
33HQJ AKQJT9 AK AK2c11224
37Ax AKQJ????? A A11c1 × 9c51216632
38Ax AKQJT??? AK A11c1 × 8c32414784
39Ax AKQJT9? AKQ A11c1 × 7c1241848
40Ax AKQJT9? AK AK11c1 × 7c112924
41Ax AKQJT9 AKQJ A11c124264
42Ax AKQJT9 AKQ AK11c124264
43Ax AKQJT AKQJT A11c112132
44Ax AKQJT AKQJ AK11c124264
45Ax AKQJT AKQ AKQ11c112132
46Ax AKQJ AKQJ AKQ11c112132
47KQ AKQJT9? AK AK7c11284
Total hands guaranteed to win 12 tricks in notrump48360

Curiously, the most common hand that will guarantee 6 NT contains a nine-card suit (Case 37). This is also the longest possible suit, because a 10-carder would require three aces on the side, so the hand would either be cold for 13 tricks (excluded by the problem conditions) or have no guarantee of 12 after an ace is knocked out.

Dan Baker: With no singleton, there are 12 choices for an imperfect suit fewer than seven cards, e.g., five cards can be AKQJx without the 10 (8 choices) or HHHT9 (4 choices); or 51 choices if seven cards (7-2-2-2): AKQJTxx without the nine (21 choices) or HHHH98? (30 choices). With a singleton, you can’t establish slow tricks after that ace is knocked out, so the imperfect suit must be 2-6 cards with 13-L choices (L = length), e.g., three cards must be AKx without the queen (13 - 3 = 10 choices).

Dan Gheorghiu: Declarer must have one suit with [the lowest card] a loser (e.g, AKQx) or a sequence broken by the absence of only one card (e.g., AKJT or KQJT). In the latter case, at least doubleton stoppers (AK) are necessary in every other suit, [because the lead must be surrendered to establish one or more tricks].

With some hands 6 NT is always makable but not guaranteed. For example, holding S A-K-Q-J-9-2 H A-K-Q D A-K C A-K, after any safe lead declarer can test spades to discover they can’t be established, and without risk come down to S J-9-2 H A D A C A. The opponent must still have S 10-8-x (else just concede a spade) so an endplay looms if declarer can strip his remaining cards, but there’s the rub. Which ace do you cash first? If you pick a suit in which your opponent has two cards, you fail. And if the first ace cashed is successful, you have the same problem on the next one. Therefore, these hands require guessing the layout, which the problem conditions explicitly forbade.

Another curiosity, as well as my inspiration for this puzzle, is how close the two answers (47,340 and 48,360) happen to be. One could argue that a difference of 1020 is sizable, but in a playing field of 635+ billion bridge hands — 635,013,559,600 to be exact — it’s pretty darn close.

In case anyone is interested, the number of hands guaranteed to make 7 NT is exactly 3756 — an easier problem solved many years ago. A summary table can be seen here.

Losers anonymous

Dan Gheorghiu: Evidently the Professor was right all along with his “boson deflection theory,” as hands with the highest freakness (eight- or nine-card suit) account for about 75 percent of those guaranteed to make 6 NT. Rumors are he will be nominated for a “Fundamental Physics Prize” for applying the laws of physics to bridge. My sincere congratulations, sir!

Grant Peacock: I spent the afternoon counting the number of ways to put a loser into a suit.

Well, so did my wife 20 years ago when she took me to a menswear store.


© 2016 Richard Pavlicek